Random 5 function will ensure equal probability for all number starting from
0 to 5. Now consider an event that you will run that random 5 function
twice. This will create 25 elements in your sample space. Now make seven
groups of three elements. This means that each group will be having three
elements now. There are other 4 elements remaining. While doing the
experiment, it we encounter one of these four elements, we will perform the
experiment again. These will ensure the 1/7 probability. Proof is as
follows:

If we perform the random 5 function twice, we will end up with getting the
followings:

(0,0)   (1,0)    .......            ...... (4,0)
(0,1)   (1,1)    .......            ...... (4,1)
(0,2)   (1,2)    .......            ...... (4,2)
(0,3)   (1,3)    .......            ...... (4,3)
(0,4)   (1,4)    .......            ...... (4,4)

 So total there are 25 elements.

  Now make seven group of 3 elements ....

  {(0,0),(0,1),(0,2)} [mark it as 0], {(0,3),(0,4),(1,0)} [mark it as 1]
..... upto 7 seven groups

 We have {(4,1),(4,2),(4,3),(4,4)} elements remaining. If we encounter these
elements during experiment, we shall ignore them and continue again. These
will ensure 1/7 probability.

  For each number ( 0 to 7):

      probability :
                         3/25 + 4/25 * 3/25 + (4/25)^2 * 3/25 +
................ upto infinity
                        = 3/25{1+ 4/25 + (4/25)^2 + ............. upto
infinity}
                        = 3/25 * [1 / (1- (4/25))]
                        = 3/25 * 25/21
                        = 1/7


On 27 June 2010 16:57, sharad kumar <sharad20073...@gmail.com> wrote:

> You have a random 5 function. Generate a random 7 function using this
> random 5 function with equal probability
>
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-- 
regards,
soumya prasad ukil

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